Show that $f(t-i) \star g(t-k)=m(t-i-k)$

Question

Let $f,g\in L^1$, and let $m=f \star g$ the convolution product of $f$ and $g$, $i,k\in \mathbb Z.$ Show that $$f(t-i) \star g(t-k)=m(t-i-k)$$

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\begin{align*} f(t-i)\star g(t-i)&=f(t-i)\star g(t+i-i-k) \\&=f(t-i)\star h(t-i) \quad(h(t-i)=g(t-k)) \\ &=\int_{\mathbb R} f(t-i-s)h(s) \\&=\int_{\mathbb R} f(t-i-s)g(s+i-k)ds\\&=\int_{\mathbb R} f(t-i-k-s')g(s'+i)ds \quad (s'=s-k)\\ \end{align*} I don't know how to complete the proof ! any help please ! I know it's a stupid question but I get stuck !

Answer 1

What do you mean by $f(t-i)\star g(t-k)$ ? What is true is that \begin{align*} m(t-i-k)&=\int_{\mathbb R} f(t-i-k-s)g(s)ds \\ &=\int_{\mathbb R}f(t-i-u)g(u-k)du\\ &=(f\star \widetilde{g})(t-i), \end{align*} where $\widetilde{g}=g(\cdot-k)$.